phyf 115 tutorial solutions all

SOLUTIONS TO PROBLEMS: CHAPTER 1

P1.1 With and , we have

P1.2 For either sphere the volume is and the mass is . We divide

this equation for the larger sphere by the same equation for the smaller:

.

Then .

P1.3 Use .

(a) For He, .

(b) For Fe, .

(c) For Pb, .

(d) For we have as the mass of one mole.

(b)

P1.4 The term x has dimensions of L, a has dimensions of , and t has dimensions of T. Therefore, the equation has dimensions of

or .

The powers of L and T must be the same on each side of the equation. Therefore,

and .

Likewise, equating terms in T, we see that must equal 0. Thus, . The

value of k, a dimensionless constant, .

P1.5 (a) since the units of are , while the units of are

.

(b) since the units of are m, and is dimensionless if

is in .

P1.6 Inserting the proper units for everything except G,

.

Multiply both sides by and divide by ; the units of G are .

*P1.7 Each of the four walls has area . Together, they have area

.

*P1.8

P1.9 Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we should expect the area to be about .

Categorize: We model the lot as a perfect rectangle to use Area = Length ´ Width. Use the conversion: .

Analyze: .

Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper units of . Unit conversion is a common technique that is applied to many problems.

P1.10 From Table 1.5, the density of lead is , so we should expect our calculated value to be close to this number. This density value tells us that lead is about 11 times denser than water, which agrees with our experience that lead sinks.

Density is defined as mass per volume, in . We must convert to SI units in the

calculation.

At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our result is indeed close to the expected value. Since the last reported significant digit is not certain, the difference in the two values is probably due to measurement uncertainty and should not be a concern. One important common-sense check on density values is that objects which sink in water must have a density greater than , and objects that float must be less dense than water.

*P1.11 The weight flow rate is .

P1.12 (a) (b) (c) (d)

P1.13

also,

.

In other words, the percentages of uncertainty are cumulative. Therefore,

,

and

.

P1.14 The actual number of seconds in a year is

.

The percent error in the approximation is

.

SOLUTIONS TO PROBLEMS: CHAPTER 2SOLUTIONS TO PROBLEMS: CHAPTER 1

P2.1 (a)

(b)

(c)

(d)

(e)

P2.2 (a) Let d represent the distance between A and B. Let be the time for which the

walker has the higher speed in . Let represent the longer time

for the return trip in . Then the times are and

. The average speed is:

(b) She starts and finishes at the same point A. With total displacement = 0,

average velocity .

P2.3 (a) at , (Point A)

at , (Point B)

(b) The slope of the tangent line is found from points C and D. and

,

.FIG. P2.7

(c) The velocity is zero when x is a minimum. This is at .

P2.4 (a)

(b)

(c)

(d) FIG. P2.9

P2.5 (a) Acceleration is constant over the first ten seconds, so at the end,

.

Then so v is constant from to . And over the last five seconds the velocity changes to

.

(b) In the first ten seconds,

.

Over the next five seconds the position changes to

.

And at ,

.

P2.6 (a) At , .

At ,

so

.

(b) At all times the instantaneous velocity is

At , .

At , .

(c)

(d) At all times . (This includes both and

).

P2.7 Given when , and at , ,

:

.

P2.8 (a) , , so , so

. Thus and .

(b) At this acceleration the plane would overshoot the runway: .

P2.9 (a) : and .

(b)

P2.10 (a) : when , . Therefore,

.

(b)

P2.11 : At , and

.

If the helicopter releases a small mailbag at this time, the equation of motion of the mailbag is

.

Setting ,

.

Solving for t, (only positive values of t count), .

P2.12 (a) We require when

(b)

(c)


P3.1 (a)

(b)

measured from the +x axis.

P3.2 We have

and .

P3.3 (a) Using graphical methods, place the tail of vector B at the head of vector A. The new vector has a

magnitude of from the x-

axis.

(b) The vector difference is found by placing the negative of vector B at the head of vector A. The resultant vector has

magnitude units at an

from the + x-axis.

FIG. P3.10

P3.4 To find these vector expressions graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor. (Scale: )

(a) A + B = 5.2 m at 60°

(b) A – B = 3.0 m at 330°

(c) B – A = 3.0 m at 150°

(d) A – 2B = 5.2 m at 300°.

FIG. P3.15

P3.5 (a) See figure to the right.

(b)

FIG. P3.27

P3.6 and

:

or

;

P3.7 Let East, North,

x y300

–1750

125

(a)

(b)

P3.8 Let q represent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and form an isosceles triangle in

which the angles are , , and . The

magnitude of R is then . [Hint:

apply the law of cosines to the isosceles triangle and use the fact that .]

Again, A, –B, and form an isosceles triangle with apex angle q. Applying the law of cosines and the identity

gives the magnitude of D as .

The problem requires that .

Thus, . This gives

and .

FIG. P3.51

P3.9 (a) , ,

(b)

(c)

*P3.10 The position vector from the ground under the controller of the first airplane is

The second is at

Now the displacement from the first plane to the second is

with magnitude

.

P3.11

FIG. P3.59

P3.12 Since

,

we have

givingFIG. P3.65

or [1]

and

. [2]

Since both vectors have a magnitude of 5.00, we also have

.

From , it is seen that

.

Therefore, gives

.

Then, and Eq. [2] gives

.

Defining q as the angle between either A or B and the y axis, it is seen that

and .

The angle between A and B is then .


P4.1 (a)

(b)

(c)

(d)

(e)

(f)

P4.2 (a) From , the x-component of velocity is

and

similarly,

and .

At ,

and .

(b)

(c) The object moves in .

P4.3 (a)

(b)

P4.4 and

(a)

(b)

(c) At

P4.5 (a) The time of flight of the first snowball is the nonzero root of

The distance to your target is

.

Now the second snowball we describe by

Using we can solve

and .

(b) The second snowball is in the air for time , so you throw it after the first by

.

P4.6 (a)

Setting , and , we have

. FIG. P4.50

Solving for d yields,

or .

(b) Setting leads to and .

P4.7 (a) From Part (c), the raptor dives for undergoing displacement

197 m downward and forward.

(b)

(c) ,

FIG. P4.61

P4.8 (a) See figure to the right.

(b) The components of the 20.2 and the along the rope together constitute the centripetal acceleration:

(c) so

FIG. P4.36

P4.9 (a)

(b) See figure to the right.

(c)

FIG. P4.53

P4.10 After the string breaks the ball is a projectile, and reaches the ground at time t:

so .

Its constant horizontal speed is

so before the string breaks


P5.1 For the same force F, acting on different masses and

(a)

(b)

P5.2

P5.3 We find acceleration:

Now becomes

P5.4 reads

where represents the direction of a

at below the –x-axis

.

For the vectors to be equal, their magnitudes and their directions must be equal.

(a) counterclockwise from the x-axis

(b)

(d) so

so

(c)

P5.5

The two forces acting on the block are the normal force, n, and the weight, mg. If the block is considered to be a point mass and the x-axis is chosen to be parallel to the plane, then the free body diagram will be as shown in the figure to the right. The angle q is the angle of inclination of the plane. Applying Newton’s second law for the accelerating system (and taking the direction up the plane as the positive x direction) we have

:

:

FIG. P5.22

(a) When

(b) Starting from rest

P5.6First, consider the block moving along the horizontal. The only

force in the direction of movement is T. Thus,

(1)

Next consider the block that moves vertically. The forces on it are the tension T and its weight, 88.2 N.

We have

(2)

FIG. P5.24

Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2)

can be added to give . Then

.

P5.7, ,

(a)

and

(b)

FIG. P5.26

(c) Since ,

P5.8As the man rises steadily the pulley turns steadily and the tension

in the rope is the same on both sides of the pulley. Choose man-pulley-and-platform as the system:

The worker must pull on the rope with force . FIG. P5.29

P5.9

Adding gives so

.

FIG. P5.54

(b)

and

(c) From above, and .

P5.10 (a) Following the in-chapter Example about a block on a frictionless incline, we have

(b) The block slides distance x on the incline, with

:

after time .

(c) Now in free fall :

Only one root is physical

(d) total time

(e) The mass of the block makes no difference.

P5.11

First, we will compute the needed accelerations:

FIG. P5.33

Newton’s second law is:

(a) When , .

(b) When , .

(c) When , .

(d) When , .

P5.12, , ,

(a) :

FIG. P5.41

:

(b) ,

(c) ,

(d)

where


P6.1 (a)

(b)

thus and .

Then .

*P6.2

Fg mg 4 kg 9.8 m s2 39.2 N

sinq 1.5 m2 m

q 48.6

r 2 m cos48.6 1.32 m

Fx max mv2

r

Ta cos48.6 Tb cos48.6 4 kg 6 m s 2

1.32 m

Ta Tb 109 N

cos48.6165 N

Fy mayTa sin48.6 Tbsin48.6 39.2 N 0

Ta Tb 39.2 Nsin48.6

52.3 N

q

39.2 N

T a

T b forces

v a c

motion

FIG. P6.11

(a) To solve simultaneously, we add the equations in

Ta and

Tb:

Ta Tb Ta Tb 165 N 52.3 N

Ta 217 N

2 108 N

(b)

P6.3 , ,

(a)

(b)

FIG. P6.13(a) FIG. P6.13(b)

P6.4 Let the tension at the lowest point be T.

FIG. P6.15

P6.5 At the top of the vertical circle,

or

P6.6 (a)

At the lowest point, his seat exerts an upward force; therefore, his weight seems to increase. His apparent weight is

.

(b) At the highest point, the force of the seat on the pilot is directed down and

.

Since the plane is upside down, the seat exerts this downward force.

(c) When , then . If we vary the aircraft’s R and v such that the

above is true, then the pilot feels weightless.

P6.7 For the block to remain stationary, and

.

so .

At the point of slipping, the required centripetal force equals the maximum friction force:

or .

For the penny to remain stationary on the block:

or

and .

When the penny is about to slip on the block,

or

EMBED MSDraw \* mergeformat

FIG. P6.60

This is less than the maximum speed for the block, so the penny slips before the block starts to slip.

The maximum rotation frequency is

Max rpm .

P6.8 Standing on the inner surface of the rim, and moving with it, each person will feel a normal force exerted by the rim. This inward force causes the centripetal acceleration:

:

The period of rotation comes from :

so the frequency of rotation is .

P6.9 (a) If the car is about to slip down the incline, f is directed up the incline.

where gives

and .

Then, yields

.

When the car is about to slip up the incline, f is directed down the incline. Then,

with yields

and .

In this case, , which

gives

.

(b) If , then .

(c)

FIG. P6.67

P6.10 (a) The bead moves in a circle with radius at a speed of

The normal force has an inward radial component of and an upward component of

or

FIG. P6.68(a)


P7.1 The component of force along the direction of motion is

.

The work done by this force is

.

P7.2 (a)

(b)

(c)

P7.3

P7.4

and W equals the area under the Force-Displacement curve

(a) For the region ,

(b) For the region ,

(c) For the region ,

(d) For the region

FIG. P7.13

P7.5 (a) Spring constant is given by

(b) Work

P7.6 (a)

(b) :

(c)

P7.7

(a)

(b)

P7.8 (a)

(b)

(c)

(d)

P7.9 (a) , but because he moves at constant speed. The skier rises a

vertical distance of . Thus,

.

(b) The time to travel 60.0 m at a constant speed of 2.00 m/s is 30.0 s. Thus,

.

P7.10 (a) The distance moved upward in the first 3.00 s is

.

The motor and the earth’s gravity do work on the elevator car:

Also, so .

(b) When moving upward at constant speed the applied force equals

the . Therefore,

.


P8.1 From leaving ground to the highest point,

The mass makes no difference:

P8.2 Using conservation of energy for the system of the Earth and the two objects

(a)

(b) Now we apply conservation of energy for the system of the 3.00 kg object and the Earth during the time interval between the instant when the string goes slack and the instant at which the 3.00 kg object reaches its highest position in its free fall. FIG. P8.13

P8.3 (a)

Similarly,

(b)

FIG. P8.24

P8.4 (a)

The result does not depend on the path since the force is conservative.

(b)

so

(c)

P8.5 We shall take the zero level of gravitational potential energy to be at the lowest level reached by the diver under the water, and consider the energy change from when the diver started to fall until he came to rest.

P8.6 :

FIG. P8.31

P8.7 (a)

(b)

(c) The mechanical energy converted due to friction is 86.5 J

(d)

FIG. P8.33

P8.8 (a)

(b)

(c) It’s possible to find an effective coefficient of friction, but not the actual value of m since n and f vary with position.

P8.9

FIG. P8.57

P8.10

FIG. P8.59


P9.1 (a) At maximum height , so .

(b) Its original kinetic energy is its constant total energy,

.

At the top all of this energy is gravitational. Halfway up, one-half of it is gravitational and the other half is kinetic:

Then

.

P9.2 (a) For the system of two blocks ,

or

Therefore,

Solving gives (motion toward

the left).

(b) FIG. P9.4

P9.3 (a) area under curve

(b)

(c) From the graph, we see that FIG. P9.7

*P9.4 The impact speed is given by . The rebound speed is given by

. The impulse of the floor is the change in momentum,

P9.5 Take x-axis toward the pitcher

(a) :

:

(b)

P9.6 Momentum is conserved

P9.7 (a), (b) Let and be the velocity of the girl and the plank relative to the ice surface. Then we may say that is the velocity of the girl relative to the plank, so that

(1)

But also we must have , since total momentum of the girl-plank system is zero relative to the ice surface. Therefore

, or

Putting this into the equation (1) above gives

or

Then

FIG. P9.21

P9.8 Energy is conserved for the bob-Earth system between bottom and top of swing. At the top the stiff rod is in compression and the bob nearly at rest.

:

so

Momentum of the bob-bullet system is conserved in the collision:

FIG. P9.24

P9.9 (a) First, we conserve momentum for the system of two football players in the x direction (the direction of travel of the fullback).

where q is the angle between the direction of the final velocity V and the x axis. We find

(1)

Now consider conservation of momentum of the system in the y direction (the direction of travel of the opponent).

which gives, (2)

Divide equation (2) by (1)

From which

Then, either (1) or (2) gives

(b)

Thus, the kinetic energy lost is

P9.10 By conservation of momentum for the system of the two billiard balls (with all masses equal),

FIG. P9.33

P9.11 The x-coordinate of the center of mass is

and the y-coordinate of the center of mass is

P9.12 Let represent the area of the bottom row of

squares, the middle square, and the top pair.

, , ,

FIG. P9.41


P10.1 (a)

(b)

P10.2 ,

(a)

(b)

P10.3 and are two equations in two unknowns and a

:

:

P10.4 Given , , and

(a)

At ,

(b)

The magnitude of the total acceleration is:

The direction of the total acceleration vector makes an angle f with respect to the radius to point P:

(c)

P10.5 (a)

(b)

P10.6 , ;

, ;

, ;

about the x-axis

(a)

FIG. P10.20

(b)

P10.7 Treat the tire as consisting of three parts. The two sidewalls are each treated as a hollow cylinder of inner radius 16.5 cm, outer radius 30.5 cm, and height 0.635 cm. The

tread region is treated as a hollow cylinder of inner radius 30.5 cm, outer radius 33.0 cm, and height 20.0 cm.

Use for the moment of inertia of a hollow cylinder.

Sidewall:

Tread:

Entire Tire:

P10.8 Every particle in the door could be slid straight down into a high-density rod across its bottom, without changing the particle’s distance from the rotation axis of the door. Thus, a rod 0.870 m long with mass 23.0 kg, pivoted about one end, has the same rotational inertia as the door:

.

The data.

P10.9

The thirty-degree angle is unnecessary information.

FIG. P10.31

P10.10

The magnitude of the torque is given by , where f is the force of friction.

Therefore, and

yields

FIG. P10.38

P10.11 (a) For the counterweight,

becomes:

For the reel reads

where

We substitute to eliminate the acceleration:

and

:

FIG. P10.45

(b) Use conservation of energy for the system of the object, the reel, and the Earth:

:

P10.12 where since no slipping.

Also, , , and

Therefore,

Thus,

For a disk,

So or

For a ring, so or

Since , reaches the bottom first


P11.1

(a)

(b)

or 168°

(c) Only gives the angle between the vectors unambiguously.

P11.2

, and

FIG. P11.11

P11.3

so

and

P11.4 (a) The net torque on the counterweight-cord-spool system is:

.

(b)

(c)

P11.5 (a)

(b) At the highest point of the trajectory,

and

FIG. P11.17

(c)

(d) The downward force of gravity exerts a torque in the –z direction.

P11.6 (a) Let mass of rod and mass of each bead. From , we have

When , , , and with other values as stated in the problem, we find

.

(b) Since there is no external torque on the rod,

constant and .

P11.7 (a)

FIG. P11.45

(b) Think of the whole weight, 3mg, acting at the center of gravity.

(c)

(d)

The angular acceleration is not constant, but energy is.

(e) maximum kinetic energy

(f)

(g)

(h)

P11.8 (a)

Angular momentum is conserved.

(b)

(c) The work is done by the centripetal force in the negative direction.

Method 1:

Method 2:

(d) Using the data given, we find

P11.9 (a)

(b)

FIG. P11.51

(c) Angular momentum is conserved:

(d)

(e)

(f)

P11. 10 (a)

(b)

(c)

(d)

(e)

(f)

FIG. P11.52


P12.1 Take torques about P.

We want to find x for which .

FIG. P12.3

P12.2 The coordinates of the center of gravity of piece 1 are

and .

The coordinates for piece 2 are

and .

The area of each piece is

and .

And the mass of each piece is proportional to the area. Thus,

FIG. P12.5

and

P12.3 Let the fourth mass (8.00 kg) be placed at (x, y), then

Similarly,

P12.4 Relative to the hinge end of the bridge, the cable is attached

horizontally out a distance and

vertically down a distance . The

cable then makes the following angle with the horizontal:

.

(a) Take torques about the hinge end of the bridge:

which yields

(b)

or

(c)

Thus,

FIG. P12.20

P12.5 (a)

Taking torques about an axis at the foot of the ladder,

Solving the torque equation,

.

Next substitute this value into the equation to find

FIG. P12.13

in the positive x direction.

Solving the equation ,

in the positive y direction.

(b) In this case, the torque equation gives:

or .

Since and , we find

.

P12.6 (a) Consider the torques about an axis perpendicular to the page and through the left end of the horizontal beam.

,

giving . FIG. P12.12

(b) From , , or .

From , , or .

P12.7 When , the rod is on the verge of slipping, so

.

From , , or .

Thus, FIG. P12.23

From , , or , giving .

Using for an axis perpendicular to the page and through the left end of the

beam gives , which reduces to

.

P12.8 (a) See the diagram.

(b) If , then

Solving for the tension gives: .

FIG. P12.43

From , .

From , .

(c) If :

.

Solving for x gives: .

P12.9

If the CM of the two bricks does not lie over the edge, then the bricks balance.

If the lower brick is placed over the edge, then

the second brick may be placed so that its end

protrudes over the edge. FIG. P12.24

P12.10 Using , choosing the origin at the

left end of the beam, we have (neglecting the weight of the beam)

and .

Solving these equations, we find:

(a)

(b)

FIG. P12.45

P12.11 Count the wires. If they are wrapped together so that all support nearly equal stress, the number should be

.

Since cross-sectional area is proportional to diameter squared, the diameter of the cable will be

.

P12.12


P13.1

Thus,

or

giving . The answer and is

physically equivalent.

P13.2 The force exerted on the 4.00-kg mass by the 2.00-kg mass is directed upward and given by

The force exerted on the 4.00-kg mass by the 6.00-kg mass is directed to the left

FIG. P13.5

Therefore, the resultant force on the 4.00-kg mass is

P13.3

If

then .

P13.4 (a) At the zero-total field point,

so

(b) At this distance the acceleration due to the Earth’s gravity is

P13.5

so or

FIG. P13.23

P13.6

or FIG. P13.25

P13.7 The height attained is not small compared to the radius of the Earth, so does

not apply; does. From launch to apogee at height h,

:

P13.8 (a)

(b) of the equilateral triangle

P13.9 Applying Newton’s 2nd Law, yields for each star:

or .

We can write r in terms of the period, T, by considering the time and distance of one complete cycle. The distance traveled in one orbit is the circumference of the stars’ common orbit, so

. ThereforeFIG. P13.13

P13.10 For both circular orbits,

:

FIG. P13.60(a) The original speed is

.

(b) The final speed is .

The energy of the satellite-Earth system is

(c) Originally .

(d) Finally .

(e) Thus the object speeds up as it spirals down to the planet. The loss of gravitational energy is so large that the total energy decreases by

.

(f) The only forces on the object are the backward force of air resistance R, comparatively very small in magnitude, and the force of gravity. Because the spiral path of the satellite is not perpendicular to the gravitational force,

pulls forward on the satellite to do

positive work and make its speed increase.

P13.11 (a) The work must provide the increase in gravitational energy

(b) In a circular orbit, gravity supplies the centripetal force:

Then,

So, additional work kinetic energy required

P13.12 (a) The gravitational force exerted on by the Earth (mass ) accelerates

according to: . The equal magnitude force exerted on the Earth

by produces negligible acceleration of the Earth. The acceleration of relative approach is then

.

(b) Again, accelerates toward the center of mass with . Now the

Earth accelerates toward with an acceleration given as

The distance between the masses closes with relative acceleration of

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